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Draw a Cantor set C on the circle and consider the set A of all the chords between points of C. Prove that A is compact.

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    $\begingroup$ Okay. I drew the Cantor set on the circle. What next? $\endgroup$ – Asaf Karagila Mar 16 '13 at 0:50
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    $\begingroup$ If $A \subset \mathbb R^{2}$, then we have to show $A$ is closed and bounded. Clearly $A$ is bounded, so we just have to show $A$ is closed. $\endgroup$ – Dylan Yott Mar 16 '13 at 0:55
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    $\begingroup$ @David : Why not make your comment an answer? $\endgroup$ – Stefan Smith Mar 16 '13 at 1:32
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    $\begingroup$ OK.${}{}{}{}{}$ $\endgroup$ – David Moews Mar 16 '13 at 1:51
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$C$ is compact as it's closed and bounded. Then, $A$ is compact as it's the image of the compact set $C\times C\times [0,1]$ under the continuous map $\phi: {\Bbb R}^2\times {\Bbb R}^2\times [0,1]\to {\Bbb R}^2$ given by $\phi(x,y,\lambda)= \lambda x + (1-\lambda )y$.

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  • $\begingroup$ Is this set convex? $\endgroup$ – MathCosmo Dec 1 '18 at 5:14

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